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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Just rattled off a page about the Intermediate Value Theorem

I have attempted to simplify it, and also included some examples which I may have got wrong, so I would appreciate a critical review before I let a wider audience know about the page.

Thanks!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Anyone care to tear it to pieces?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,186

Hi MathsisFun;

MathsisFun wrote:

Anyone care to tear it to pieces?

Rather different tone than normal. Almost a challenge. You are proud of this one. You should be.

Very interesting page. You cleared up a point or two for me. I didn't know that about the table problem. Here is a link to the proof.

http://arxiv.org/abs/math/0511490

I guess I can only say the biggest compliment Is that I saved the page for my notebook.

*Last edited by bobbym (2009-11-04 22:04:59)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Wow, thanks!

But the "round trip" example was my own invention and is thus likely to be wrong.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'd say the round trip better demonstrates Rollé's Theorem instead of the IVT.

ie. You start and finish at the same height, so one part of your route must be flat.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Rolles theorem requires the curve to be differentiable (i.e. smooth) not just continuous. MathsIsFuns example is about continuous routes, not smooth ones. You could e.g. walk up the side of a pyramid to the apex and down the another side in which case no part of your route is horizontal.

*Last edited by JaneFairfax (2009-11-05 05:36:43)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Or your route could be like this:

[align=center]http://www.mathsisfun.com/graph/functio … =-2&ymax=2[/align]

In which case the IVT applies but Rolles theorem does not!

*Last edited by JaneFairfax (2009-11-05 05:37:02)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The IVT says a lot more than "you will be at the same height". If you walk in a circle (this works for any parametrized continuous path, but circle is easiest to see), let your height at time t be described by h(t). Here we are walking around the circle in 1 unit time of time. Then there will exist at least one point which will be exactly the same height as the opposite side of the circle.

We can define a function:

g(t) = h(t) - h(t+1/2)

If g(0) = 0, then we're done: our starting point is at exactly the same height as the opposite side. Otherwise, g(0) is either positive or negative. But g(1/2) has exactly the opposite sign of g(0):

g(0) = h(0) - h(1/2)

g(1/2) = h(1/2) - h(0)

(Remember that h(0) = h(1). They are exactly the same point!)

Now the IVT says that g(t) must be zero at some point in between 0 and 1/2. And that's it.

Of course, probably too difficult to put in the page...

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

It would be nice to include that!

Maybe "if you travel in a circle, then *at some point* the height opposite you on the circle will be exactly the same as where you are" ... or some such wording.

Or "when you have a circle with a wavy circumference, somewhere there will be two points opposite one another that are exactly the same height".

A rather-difficult 3d sketch could complement that.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,186

Hi all;

MathsisFun wrote:

But the "round trip" example was my own invention and is thus likely to be wrong.

We could argue for awhile about why it is "likely to be wrong" just because it is your own invention. True, that example does have some kinks in it, but does that mean there was a greater than 50% chance that it was an error from the start.

I thought the table problem was much more interesting, just because I have never seen even the handiest people level a table like that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

bobbym wrote:

True, that example does have some kinks in it, but does that mean there was a greater than 50% chance that it was an error from the start.

Kinks? What Kinks?

Ricky wrote:

The IVT says a lot more than "you will be at the same height". If you walk in a circle (this works for any parametrized continuous path, but circle is easiest to see), let your height at time t be described by h(t). Here we are walking around the circle in 1 unit time of time. Then there will exist at least one point which will be exactly the same height as the opposite side of the circle.

I have attempted to illustrate that ... please have a look at the revised page (at bottom).

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I have attempted to illustrate that ... please have a look at the revised page (at bottom).

Looks good, I suppose an attempt at proof would be just a little too much? It is rather surprising how elementary the proof is.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,186

Hi Mathsisfun;

Kinks? What Kinks?

Correction, No kinks. Only, I haven't seen the IVT for only a 2D graph and not a 3D surface. Following up on it I was led to the Borsuk - Ulam theorem which may or may not apply.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Denominator****Member**- Registered: 2009-11-23
- Posts: 155

The intermediate value theorem sounds easy!

How do you measure the length from a to b?

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